Solving Trigonometric Equations

We are given two trigonometric equations and conditions on the angles x and y:

• $\cos(x + y) = \frac{1}{2}$
• $\sin(x - y) = 0$
• x and y are positive acute angles ($0^\circ < x < 90^\circ$, $0^\circ < y < 90^\circ$)
• $x \ge y$

Finding Angle Values

First, consider the equation $\cos(x + y) = \frac{1}{2}$.

• Since x and y are positive acute angles, their sum $x + y$ must be between $0^\circ$ and $180^\circ$ ($0^\circ < x + y < 180^\circ$).
• The angle in this range whose cosine is $\frac{1}{2}$ is $60^\circ$.
• Therefore, we have the equation:

  $x + y = 60^\circ \quad (1)$

Next, consider the equation $\sin(x - y) = 0$.

• Given that $x \ge y$ and both are positive acute angles, the difference $x - y$ must be between $0^\circ$ and $90^\circ$ ($0^\circ \le x - y < 90^\circ$).
• The angle in this range whose sine is $0$ is $0^\circ$.
• Therefore, we have the equation:

  $x - y = 0^\circ \quad (2)$

Calculating x and y

Now we solve the system of linear equations formed by (1) and (2):

1. Add equation (1) and equation (2):

   $(x + y) + (x - y) = 60^\circ + 0^\circ$

   $2x = 60^\circ$

   $x = 30^\circ$

2. Substitute the value of x back into equation (1):

   $30^\circ + y = 60^\circ$

   $y = 60^\circ - 30^\circ$

   $y = 30^\circ$

We verify that $x = 30^\circ$ and $y = 30^\circ$ satisfy the conditions: they are positive acute angles, and $x \ge y$.

Thus, x and y are $30^\circ$ and $30^\circ$.